In the case of electrons and photons having the same wavelength, what is the same for them?
A. Energy
B. Velocity
C. Momentum
D. Angular momentum

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Hint: Equate the de Broglie wavelength of both electrons and protons and solve the question.

Complete step by step solution:
Given, the wavelength of electrons and photons are the same.
Now, the wavelength of electrons are given by \[{\lambda _e}\] and the wavelength of protons are given by \[{\lambda _p}\].
Thus , the wavelength of electrons = the wavelength of protons
\[{\lambda _e} = {\lambda _p}\]
From de Broglie wavelength, we know that \[\lambda = \dfrac{h}{p}\]
  \dfrac{h}{{{p_e}}} = \dfrac{h}{{{p_p}}} \\
   \Rightarrow \dfrac{1}{{{p_e}}} = \dfrac{1}{{{p_p}}} \\
   \Rightarrow {p_e} = {p_p} \\

Thus , the velocity of electrons and protons will be the same .

Hence, the required option is C - momentum.

Additional information:
While solving the sum the student needs to be acquainted with the de Broglie wavelength and the formulae. When solving the sum, the constant gets cancelled out from both the sides leaving behind the momentum of both the particles. Thus, it can be said that the momentum of the electrons and protons are equal when the wavelength of both the particles are equal. Students usually go wrong deciding which formula to use and thus how to solve the sum.

Note: From de Broglie wavelength , we know that the wavelength of a microscopic particle can be given by \[\lambda = \dfrac{h}{p}\] where \[\lambda \] is the wavelength of given particle, h is Planck’s constant which is equal to \[6.626 \times {10^{ - 34}}{m^2}kg/s\] and p is momentum of the given particle. Before solving the question, it needs to be understood conceptually.